theo tính chất dãy tỉ số bằng nhau ta có: \(\frac{\left(7x-5\right)+\left(6x-4\right)}{6+4}=7x+6y-9\Leftrightarrow\frac{7x+6y-9}{10}=7x+6y-9\Leftrightarrow63x+54y-81=0\)
lại có: \(\frac{7x-5}{6}=\frac{6y-4}{4}\Rightarrow28x-20=36y-24\Rightarrow7x=9y-1\)
nên \(63x+54y-81=0\Leftrightarrow7x\cdot9+54y-81=0\Leftrightarrow9\left(9y-1\right)+54y-81=0\Leftrightarrow81y-9+54y-81=0\Leftrightarrow135y-90=0\Leftrightarrow y=\frac{90}{135}=\frac{2}{3}\Rightarrow x=\frac{9y-1}{7}=\frac{5}{7}\)